# Tag Info

24

Hint: $|\sqrt2 -1|<1/2$, so as $n\to\infty$ we have that $(\sqrt2-1)^n\to ?$ In addition to that use the fact that the set $S$ is a ring, i.e. closed under multiplication and addition.

23

This is problem 11162, posed by Paolo Perfetti, in the June-July 2005 issue of the American Mathematical Monthly. The solution below, due to the Microsoft Research Problems Group, is found in the February 2007 issue of the same magazine. For positive integer $n$, define $$A_n=[0,2^n)\cap\{k\in \mathbb{N}:|\!\sin k|<\textstyle{1\over n}\},\quad ... 18 Supose not, so that there exists an \varepsilon>0 such that (0,\varepsilon)\cap S=\emptyset. \qquad\qquad\qquad(\star) It follows that \alpha=\inf S\cap(0,+\infty) is a positive number. The choice of \alpha and its positivity implies that the one and only element of S which is in [0,\alpha) is 0. I claim that \alpha\in S. ... 18 Hint: \pi is not a Liouville number, so there exists m\in\mathbb{N} such that for all p,q\in\mathbb{Z} with q>1, we have$$ \left| \pi - \frac{p}{q}\right| \geq \frac{1}{q^m}.$$This should allow you to keep \sin n away from 0. Edit: Full Solution: Let m be as above. So for all p,q\in\mathbb{Z} with q>1 we have$$ \left| \pi - ...

17

Consider $x = \sum_{j=1}^\infty a_j/4^j$ where each $a_j$ is either $1$ or $2$. Thus the binary expansion of $x$ consists of two-digit blocks which are either $10$ or $01$. Then $x$ is far. But there are uncountably many choices, so all but countably many of them are irrational.

16

claim. The series $\sum\limits_{n=1}^\infty \sin^n(n)$ diverges. Lemma. For all number $x$ irrational there exist a rational sequence $\{\frac{p_n}{q_n}\}$ where $\{q_n\}$ is odd such that $$\left\vert x-\frac{p_n}{q_n}\right\vert<\frac{1}{q_n^2}$$ Proof. Define $x_n=\frac{1}{x_{n-1}-\lfloor x_{n-1}\rfloor}$. Let $a_0=\lfloor x\rfloor$ and ...

16

This isn't exceptionally good compared to the partial convergents of the continued fraction expansion. Terminating the continued fraction for $\pi$ right before the $292$ gives $\frac{355}{113}= \textbf{3.141592}9203\ldots$, which gets $6$ digits after the decimal place right, while your fraction only gets $3$ digits after the decimal place correct even ...

11

It converges to $0$, in fact $\frac{1}{k^7\sin{k}}$ already converges to $0$, see Theorem 2 here. This theorem gives a nice characterization of the irrationality measure of $\pi$ as the borderline number $\mu$ such that $\frac{1}{k^{u-1}\sin{k}}$ converges to $0$ for $u>\mu$, and diverges for $u<\mu$. So $\frac{1}{k^7\sin{k}}$ converges because $\mu$ ...

11

Using the property stated in that article: $$\mu(x)=2 + \limsup \frac{\log a_{n+1}}{\log q_n}$$ where the continued fraction expansion for $x$ is $[a_0,a_1,...]$ and the $n$th convergent is $\frac{p_n}{q_n}$. Start with $a_0=2$ and $a_1=2$, so $q_0=1$, $q_1=2$. Now, assume you have a continued fraction $$\frac{p_n}{q_n}=[a_0,...,a_n]$$ Define $a_{n+1}$ ...

10

Though it is clear what would count as a positive answer, the question is not precise enough that to make clear what counts as a negative answer. Nonetheless, I believe that no similar formula could possibly be equivalent to denseness. The problem is that there are sequences that are dense but that are wildly non-equidistributed. For example, your formula ...

9

Actually, we can do it for higher powers as well. Let $F_k (N)$ denote the sum of the $k^{th}$ powers of the differences in the Farey sequence. Then it is easy to see that $$F_0(N)=\sum_{n=1}^N \phi(n)\sim \frac{3N^2}{\pi^2}$$ and $$F_1(N)=1.$$ It seems you are looking for $F_2 (N)$, and, you are correct that the asymptotic is $\frac{\log N}{N^2}$. More ...

9

There are some details that I haven't fully vetted but here's a long sketch which I believe should show divergence. Define $c_n := \frac{n}{2\pi} \pmod 1$ and let $[a,b]$ be any interval in the torus $\mathbb R/\mathbb Z$. The discrepancy $D(N)$ of the sequence $c_n$ is defined to be the difference between $\# \{ n \le N : c_n \in [a,b]\}$ and the expected ...

8

I proceed the same way as in my other answer here http://math.stackexchange.com/a/110019/17445 $|\sin n| \le \varepsilon$ implies that there exists an integer $k(n)$ such that $n = k(n)\pi + a(n)$ where $a(n) \in [-\arcsin(\varepsilon) ; \arcsin(\varepsilon)] \subset [-\pi \varepsilon/2 ; \pi \varepsilon/2]$, and if both $|\sin n|$ and $|\sin m|$ are less ...

8

Suppose that the prime factors of $n^2+1$ are all bounded by $N$ for infinitely many $n$. Then infinitely many integers $n^2 + 1$ can be written in the form $D y^3$ for one of finitely many $D$. Explicitly, the set of $D$ can be taken to be the finitely many integers whose prime divisors are all less than $N$, and whose exponents are at most $2$. For ...

8

normal $\; \not\Rightarrow \;$ irrationality measure $2$ There exist numbers that are normal with irrationality measure $>2$. In fact, there exist normal numbers (meaning normal with respect to every base) that have irrationality measure $\infty.$ This is Theorem 2 in Bugeaud [2] (2002). For related results, see [1] and [3]. irrationality measure $2$ ...

7

The problem is that equidistribution is a property of sequences, but density is a property of sets. You may wish to prove as an exercise that a countable subset of $[0,1]$ is dense if and only if it can be ordered in such a way as to be equidistributed. So the set $\lbrace\,x_0,x_1,\dots\,\rbrace$ is dense if and only if there is a permutation $\sigma$ such ...

7

Assume that $a, b \in \mathbb N$ are not perfect squares and $\sqrt a + \sqrt b = n \in \mathbb N$. Then $$a = (n - \sqrt b)^2 = n^2 - 2 n \sqrt b + b$$ which means that $\sqrt b$ is a rational number. This contradicts the fact that the square root of an integer is either an integer or a irrational number.

7

Let $x$ be a fixed positive real. Then, the behavior of the sequence $\{ kx \}$ falls under one of the following three types, depending on $x$. If $x$ is an integer: In this case, the sequence $\{ kx \}$ is the constant sequence $0$. In particular, the sequence converges to $0$. If $x$ is rational, but not an integer: Suppose $x = a/b$ where $a,b$ are ...

7

Suppose $a+b=c$, so that $a+b-c=0$, with $a^3, b^3, c$ all rational. Then we have $-3abc=a^3+b^3-c^3$ by virtue of the identity $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ (take $c$ with a negative sign) Hence $a+b$ and $ab$ are both rational, so $a$ and $b$ satisfy a quadratic equation with rational coefficients. There are lots of ways of ...

6

When I first came across your question, I thought it was a modern-day approximation by somebody using a computer. But when d125q pointed out it was by Ramanujan, then I figured out he must have used a systematic method. One way is to use a Ramanujan-Sato pi formula like, $$\frac{1}{\pi} = \frac{1}{16}\sum_{n=0}^\infty ... 6 Taking a step into generalization, it is true that every additive subgroup G of \mathbb R is either discrete or dense. This can be proved by considering \alpha = \inf \{ x \in G : x>0 \}. Then G is discrete iff \alpha >0, in which case G=\alpha \mathbb Z. In your case, Jyrki's suggestion implies that \alpha=0 and so S is dense. 6 You can determine this by looking at the continued fractions of 0.8575 and 0.8585. They are [0, 1, 6, 57] and [0, 1, 6, 14, 1, 8, 2] so anything in [0, 1, 6, 15..57] would work. [0, 1, 6, 15] has the smallest denominator in that range which corresponds to 0+1/(1+1/(6+1/15)) = 91/106. I just tested smaller denominators and this checks out right. ... 5 If we look at the continued fraction of the expression \frac{\pi}{180}, we get:$$\frac\pi{180} = [0,57,3,2,1,1,1,2,40,\dots]$$And:$$\frac{44}{2521} = [0,57,3,2,1,1,1,1]$$So that's a pretty good match. It is between two terms of the continued fraction expansion of \frac\pi{180}, in particular. Now, whether this is surprisingly good in any way, ... 5 The following shows that for n=3 the ratio can be as close to 1 as we like. For any positive integer m let: a = 3m^3 + 3m^2 + 2m + 1 b = 3m^3 + 3m^2 + 2m c = 3m^2 + 2m + 1 d = m Then it is readily shown that a^3 = b^3 + c^3 + d^3. If we focus on the 'wrong solution' a^3 = b^3 + c^3, the ratio of LHS to RHS is: ... 5 Let a\sqrt{2} = m + \alpha where m is an integer and 0 < \alpha < 1, and b\sqrt{7} = n + \beta where n is an integer and 0 < \beta < 1. (The fractional parts are positive since \sqrt{2} and \sqrt{7} are irrational.) We are to prove that$$ (m+ \alpha)(n + \beta) - mn \geq 1.$$We have$$\alpha = a\sqrt{2} - m = \frac{2a^2 - ...

5

The Weyl equidistribution theorem says that for irrational $\alpha$ and sufficiently many $k$, the fractional parts $\{\alpha k\}$ will be equidistributed in the interval $[0,1]$. To apply this to your particular problem, consider a large interval $k\in[N,2N]$. Because of equidistribution, the numbers $k$ will "hit" the condition $\{\alpha k\} < ... 5 Proof sketch. First write $$\left|\frac{p}{q}-\sqrt{2}\right| = \frac{|p^2-2q^2|}{q^2\left|\frac{p}{q} + \sqrt{2}\right|}$$ The Pell equation $$p^2 - 2q^2 = 1$$ is known to have infinitely many solutions (should be proven) so $$\left|\frac{p}{q}-\sqrt{2}\right| = \frac{|p^2-2q^2|}{q^2\left|\frac{p}{q} + \sqrt{2}\right|} = ... 5 If the limit exists, then it should be equal to 1, because |\sin n| is dense in [0,1] and there exists a subsequence |\sin n_k| converging to 1. Then$$\sqrt[n_k]{|\sin n_k|} \to 1.$$5 A proof of the fact that the set of reals with irrationality measure$\gt 2$has Lebesgue measure$0\$ can be found in several places. I think the result is due to Khinchin, and is in his book on continued fractions. Here is an online proof by Beukers. It is towards the end of the paper, but independent of the rest, which anyway is worth reading. It ...

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